The Student Room Group
Reply 1
y = 3xsin2x
y' = 3sin2x + 6xcos2x
y'' = 6cos2x + 6cos2x - 12xsin2x
y'' = 12cos2x - 12sin2x

y'' + 4y = 12cos2x - 12sin2x + 12xsin2x
y'' + 4y = 12cos2x = kcos2x
k = 12
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Reply 2
porter2002
im struggling with a question here:
given that 3xsin2x is a particular integral of the differentail equation d2y by dx^2 + 4y = kcos2x, where k is a constant, find the value of k

i found d2y by dx^2 to equal (1-12x)sin2x + 12cos2x

adding 4y gives 12cos2x + sin2x

therefore
12cos2x + sin2x = kcos2x

which is where im stuck, how do i find k, the constant? :confused:

y=3xsin2xy=3xsin2x
dydx=3sin2x+6xcos2x\frac{dy}{dx} = 3sin2x + 6xcos2x
d2ydx2=6cos2x+6cos2x12xsin2x=12cos2x12xsin2x\frac{d^2y}{dx^2} = 6cos2x + 6cos2x -12xsin2x = 12cos2x -12xsin2x
12cos2x12xsin2x+4(3xsin2x)=k(cos2x)12cos2x -12xsin2x + 4(3xsin2x) = k(cos2x)
12cos2x=k(cos2x)12cos2x =k(cos2x)
12=k12=k

Hope that helps.
Reply 3
You just made a mistake differentiating :smile:

first of all differentiate the particular integral y=3xsin2x to get:

dy/dx=6xcos2x+3sin2x

then differentiate again to get d^2y/dx^2 =-12xsin2x+6cos2x+6cos2x
= 12(cos2x-xsinx)

then d^2y/dx^2 + 4y = 12(cos2x-xsinx)+12xsin2x
= 12cos2x

this must be identically equal to kcos2x for all x so k=12

hope that helps and ive differentiated correctly :smile:
Reply 4
joe8232
y=3xsin2xy=3xsin2x
dydx=3sin2x+6xcos2x\frac{dy}{dx} = 3sin2x + 6xcos2x
d2ydx2=6cos2x+6cos2x12xsin2x=12cos2x12xsin2x\frac{d^2y}{dx^2} = 6cos2x + 6cos2x -12xsin2x = 12cos2x -12xsin2x
12cos2x12xsin2x+4(3xsin2x)=k(cos2x)12cos2x -12xsin2x + 4(3xsin2x) = k(cos2x)
12cos2x=k(cos2x)12cos2x =k(cos2x)
12=k12=k

Hope that helps.

okay, i understand the above, although i have an extra sin2x from somewhere? :confused:

i think its from my working out of y'',
withing my working for y''

i got y' of 3sin2x to = sin2x, is this wrong?
Reply 5
y' of 3sin2x should be 3(sin2x)' = 3(2cos2x) by chain rule and differentiating sine, which then equals 6cos2x